JEE MAIN - Mathematics (2023 - 10th April Morning Shift - No. 18)
The number of elements in the set $$\{ n \in Z:|{n^2} - 10n + 19| < 6\} $$ is _________.
Answer
6
Explanation
Given, $\left|n^2-10 n+19\right|<6$
$\Rightarrow-6 < n^2-10 n+19 < 6$
Take, $-6 < n^2-10 n+19$ and $n^2-10 n+19 < 6$
$$ \begin{array}{ll} \Rightarrow n^2-10 n+25 > 0 & \text { and }\quad n^2-10 n+13 < 0 \\\\ \Rightarrow(n-5)^2 > 0 & \text { and } n=\frac{10 \pm \sqrt{100-52}}{2}<0 \end{array} $$
$\Rightarrow n \in \mathbb{Z}-\{5\}$
$$ \begin{array}{lr} & \therefore n \in[5-2 \sqrt{3}, 5+2 \sqrt{3}] \\\\ & \therefore n \in[13,8.3] \\\\ & \therefore n=2,3,4,5,6,7,8 \end{array} $$
Thus, number of element in the set is ' 6 '
$\Rightarrow-6 < n^2-10 n+19 < 6$
Take, $-6 < n^2-10 n+19$ and $n^2-10 n+19 < 6$
$$ \begin{array}{ll} \Rightarrow n^2-10 n+25 > 0 & \text { and }\quad n^2-10 n+13 < 0 \\\\ \Rightarrow(n-5)^2 > 0 & \text { and } n=\frac{10 \pm \sqrt{100-52}}{2}<0 \end{array} $$
$\Rightarrow n \in \mathbb{Z}-\{5\}$
$$ \begin{array}{lr} & \therefore n \in[5-2 \sqrt{3}, 5+2 \sqrt{3}] \\\\ & \therefore n \in[13,8.3] \\\\ & \therefore n=2,3,4,5,6,7,8 \end{array} $$
Thus, number of element in the set is ' 6 '
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